JEE 2023 — Mathematics PYQ

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Question

JEE 2023 — Mathematics PYQ

JEE | Mathematics | 2023

Consider a circle $C_{1}$ : $x^{2} + y^{2} - 4 x - 2 y = α - 5$ . Let its mirror image in the line $y = 2 x + 1$ be another circle $C_{2}$ : $5 x^{2} + 5 y^{2} - 10 x y - 10 x y + 36 = 0$ . Then let $r$ be the radius of $C_{2}$ . Then $r + r$ is equal to ________.

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Explanation

We have,
$C_{1} : x^{2} + y^{2} - 4 x - 2 y = α - 5$
$C_{1} : (x - 2)^{2} + (y - 1)^{2} - 5 = α - 5$
$C_{1} : (x - 2)^{2} + (y - 1)^{2} = (α)^{2}$
So, centre and radius of $C_{1}$ are $(2, 1)$ and $α$ respectively.
Now, image of $(2, 1)$ along the line $y = 2 x + 1$ is,
$\frac{x - 2}{2} = \frac{y - 1}{- 1} = \frac{- 2 ( 4 - 1 + 1 )}{2 ^{2} + ( - 1 ) ^{2}}$

$\Rightarrow \frac{x - 2}{2} = \frac{y - 1}{- 1} = \frac{- 8}{5}$

$\Rightarrow x = \frac{- 6}{5}$ and $y = \frac{1}{3} 5$

# Now, $(\frac{- 6}{5}, \frac{13}{5})$ will be the centre of C $_{2}$

$∴ f = \frac{6}{5}$ and $g = \frac{-}{1} 35$

Now, radius of $C_{2} = r = f^{2} + g^{2} - \frac{36}{5}$

$\Rightarrow r = \frac{36}{25} + \frac{169}{25} - \frac{36}{5} = 1$

$∵ r = 1 so, α = 1 ∴ α + r = 1 + 1 = 2$

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